Question

plz answer this correctly​

Answer 1

Explanation:

I give you 2 method my friend because you owned chosen your favourite method .

MY FRIEND FOR YOU THIS

ANSWER ☺️

Answer 2

Given:

$\rm 9.8\% \: w/w \: H_2SO_4$

To Find:

$\rm Molality \: of \: H_2SO_4 \: (m)$

Answer:

Let weight of solution be 100 g.

So,

Given mass of solute = 9.8 g of $\sf H_2SO_4$ is present in the solution

Mass of solvent = (100 - 9.8) g = 90.2 g of water is preset in solution

Molecular mass of $\sf H_2SO_4$ = 98 g/mol

Molality is defined as number of moles of solute present in 1 kg of solvent. Thus,

$\boxed {\boxed { \bf{Molality = \dfrac{Given \ mass \ of \ solute \ (g)}{Molecular \ mass \ of \ solute \ (g)} \times \dfrac{1000}{Mass \ of \ solvent \ (g)}}}}$

By substituting values we get:

$\rm \implies Molality = \dfrac{9.8}{98} \times \dfrac{1000}{90.2} \\ \\ \rm \implies Molality = \dfrac{1}{ \cancel{10}} \times \dfrac{100 \cancel{0}}{90.2} \\ \\ \rm \implies Molality = \dfrac{100}{90.2} \\ \\ \rm \implies Molality = 1.1 \: m \\ \\ \rm \implies Molality \approx 1 \: m$

$\therefore$ $\boxed{\mathfrak{Molality = 1 \ m}}$